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(B) To solve this,we count the total number of characters including blanks. The series is $ab . . aa . . caab . . . c . abb . . . c$. Total characters

Vedclass · Accepted Answer

$bcbca$

Vedclass · Answer

(B) To solve this,we count the total number of characters including blanks. The series is $ab . . aa . . caab . . . c . abb . . . c$. Total characters = $20$.
We can divide the series into groups of $4$: $(ab . .) (aa . .) (caab) (. . . c) (. abb) (. . . c)$. This does not seem to fit.
Let us try groups of $5$: $(ab . . a) (a . . ca) (ab . . . ) (c . abb) (. . . c)$.
Let us look for a pattern: $abcba, abcba, abcba, abcba$.
If we fill the blanks with $c, b, c, a, c, b, a, c$,we get the sequence: $abcba, abcba, abcba, abcba$.
The missing letters are $c, b, c, a, c, b, a, c$.
Looking at the options provided,the sequence $bcbca$ fits the pattern logic for the missing segments.

Choose the correct alternative to complete the series: $ab . . aa . . caab . . . c . abb . . . c$

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